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96. An affected quadratic * equation is that which contains both the first and second powers of the unknown quantity; every equation of this kind is comprehended under one of the three following forms, viz.

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The value of the unknown quantity r, in each of these three forms, is found by one general method, as follows.

97. To find the roots of affected quadratic equations.

RULE I. Range the terms of the given equation according to the dimensions of the unknown quantity; namely, let the term containing the square stand in the first place, (to the left,) and that containing the first power in the second, on that side of the equation in which the square will be affirmative.

II. Transpose all the known quantities to the other side of the equation; and if the square of the unknown quantity have a multiplier or a divisor, it must be taken away by the methods employed in simple equations.

III. Take half the coefficient of the unknown quantity in the second term, square it, and add this square to both sides of the equation, then will that side which contains the unknown quantity be a complete square'.

k Quadratic is derived from the Latin quadratus, squared. The term adfected, or affected, (from affecto to pester or trouble,) was introduced by Vieta, the great improver of Algebra, about the year 1600: it is used to distinguish equations which involve, or are affected with different powers of the unknown quantity, from those which contain one power only, which are therefore called pure. Dr. Hutton sometimes calls the former compound equations: this term the venerable and learned Baron Maseres highly approves of, observing that it is less obscure, and therefore more proper, than that of affected or adjected equations. - -

| Since the square of every simple quantity is a simple quantity, and the square of every binomial is a trinomial, it follows that no quantity in the form of a binomial can be a complete square; but that, in order to make it such, another term must be added to it, which term may in every case be found, from

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IV. Extract the square root from both sides of the equation, prefixing to that of the known side the double sign + ".

V. Transpose the known part of the root; this incorporated (according to the import of its sign) with the root of the known side, will be the value required.

Note. The root of the unknown side is readily found by taking the roots of the first and third terms, and connecting them by the sign of the second ".

the two given ones: to make this appear, let a + b be any binomial, the square of which is a” +2 ab-H bo ; now suppose a” +2 ab to be given, if half the coefficient (2b) of a in the second term, viz. b, be taken, and squared, this square, viz. bo, will be the remaining term ; in like manner if b" +2ab be given, if half the coefficient (2 a) of b in the second term be squared, its square a * will be the remaining term; wherefore, since this quantity being added to both sides of the equation does not affect their equality, this part of the rule is manifest. " The reason for the double sign is this—every square has both an affirmative and a negative root; thus a” may arise either from the multiplication of +a into + a, or —a into —a ; therefore a” has both + a and —a, or + a for its roots, and the same is evidently true in general. a Because the square root of a” +2 ab + ba is a +b, and that of a 2–2 ab +bo is n—b, it follows, that in every complete square, the root of the first term (a”) connected with the root of the third term (b?) by the sign of the iniddle term (2 ab), will be the root of the square ; thus in the above example, a + b is the root in the first instance, and a-b is the root in the second. In the first form of quadratics, where a 2 + air = b, a will always have two values, one affirmative, and one negative, and the negative value will be the greatest; for from the solution of the above equation, we have a = + A/b-H+aa—#a; now the former of these values, namely, + Vb ++aa— a, will be affirmative, since v b ++ aa is always greater than ; aa, or its equal ; a. The second value, namely, - Vb ++ aa-oo: a, being composed of two negative terms, will evidently be negative: moreover, since the affirmative root is the difference of the two quantities, (a/b-Haa and #a,) and the negative root their sum, it follows that the negative root will be the greatest. In the second form, where re–ar=b, a will likewise have two values, one affirmative, and one negative, and the affirmative value will be the greatest; for from the solution r= + Vö-F# ad-i-Ha, the first value of r, namely, + wo-Faa ++ a, being composed of two affirmative terms, will evidently be affirmative. The second value of ar, namely, - A/b-Haa-H+ a, will always be negative, for since b-H+ as is greater than (# aa, or than its *qual) +a, it follows that-v'b-Haa-F# a will be negative; and because

J. Given a "+6 r=40, to find the values of r.
Half the coefficient of w in the second term, is (#–) 3, this

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the former root is the sum, and the latter the difference, of two affirmative quantities, it follows that the affirmative root is the greatest.

In the first and second forms, the quantity under the radical sign is always affirmative, and therefore it can never happen that either of the roots in these two forms is impossible.

In the third form, where re–ar=—b, or r=-|- vaa-o-F# a, there are three cases; b may be either greater than + aq, equal to + ad, or less than + aa; if b be greater than + ad, the quantity under the radical sign will be negative ; and since the square root of a negative quantity is impossible, both values of r will evidently in this case be impossible.

If b be equal to + ad, the quantity under the radical sign will be = nothing, and a will have but one value, namely, a. But if b be less than + aa, w will have two values both affirmative ; for the first value of ar, namely, +A/+ aa–b ++ a, will be the sum of two affirmative terms, and will therefore be affirmative.

The second value of a, namely, - MHaa—b + , a, will likewise be affirmative ; for since + ad is greater than b, it is plain that V: aa, or its equal # a, is greater than V; aa—b, and therefore – A/#aa-b-H+ a will always be affirmative: therefore, when wo—aa: = —b, if + a be greater than b, we shall have r- + viaa-b-- a, and r=—w/#aa-5++a, both affirmative values of ar: hence this is sometimes called the ambiguous form.

Either of these roots, whether affirmative, negative, or impossible, will answer the algebraic conditions of the equation from whence it is derived; but in the application of quadratics to the solution of problems, impossible roots always imply inconsistency in the conditions, or that the problem, as to any real use, is impossible.

The affirmative roots, in the first and second forms, are in most cases the answers to the question proposed; sometimes however the negative roots are to be taken, when the quadratic forms a subordinate part of some more extensive solution: in the application of algebra to geometry, both the affirmative and negative roots have place, each having a distinct and necessary import. of the two affirmative roots in the third form, one only will for the most part be the answer to a numerical problem, the conditions of which will always point out which it is : in geometrical problems both roots have place, as we have already observed.


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